After years of maintaining a steady population of 32,000, the population of a town begins to grow exponentially. After 1 year an

Question

1 month ago

10

After years of maintaining a steady population of 32,000, the population of a town begins to grow exponentially. After 1 year an

d an increase of 8% per year, the population is 34,560. Which equation can be used to predict, y, the number of people living in the town after x years? (Round population values to the nearest whole number.)
y = 32,000(1.08)x
y = 32,000(0.08)x
y = 34,560(1.08)x
y = 34,560(0.08)x

Mathematics

2 answers:

Svet_ta [12.7K] · Answer 1 · 2026-02-27T19:35:39+03:00

Svet_ta [12.7K]1 month ago

4 0

y=32,000(1.08)x

Starting value: 32,000

Growth factor: 1.08, indicating an increase of 8 percent from the initial amount

Inessa [12.5K] · Answer 2 · 2026-02-27T19:36:36+03:00

Answer:

$y=32000(1+0.08)^x$

Step-by-step explanation:

The exponential growth formula is $y=a(1+r)^x$

Where 'a' signifies the initial population

r is the growth rate and x represents time in years

maintaining a constant population of 32,000. Thus, the initial population is 32,000

with an annual increase of 8%. The growth rate is 8%, equivalent to 0.08

a = 32000 and r = 0.08

Now, substituting all the values into the general formula

$y=a(1+r)^x$

$y=32000(1+0.08)^x$

$y=32000(1+0.08)^x$